! 1 ואנטגרלי אנפיניטסמלי Nחשבו

! חזרה

5/6/2010

Dedekind Cuts דדקינד!) (חתכי

A Dedekind cut (D-cut) (aka lower-D-cut) is a nonempty set A $ Qso that

(i) (−∞, a) := {q ∈ Q : q < a} ⊂ A ∀ a ∈ A;

(ii) @ maximal element in A (i.e. @ LUBA ∈ A).Let R := {lower-D-cuts} and order R by inclusion (i.e. forA, B ∈ R, A < B if A $ B), then

, (R, <) is a complete ordered set.

- Density of the rationals.

Suppose that A, B ∈ R and that A < B, then ∃ q ∈ Q so thatA < (−∞, q) < B.

Decimal representation עשרונית!) (הצגה of lower-D-cuts

Decimal representation is a map π : R→ Z×DN where

D := {digits} = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

defined by π(A) = (N; d1, d2, . . . ) where

max {a ∈ A : 10na ∈ Z} = N +n∑

k=1

dk10k∀ n ≥ 1.

• π : R→ Z× {a ∈ DN : #{k ≥ 1 : ak ≥ 1} =∞} is a setcorrespondence (bijection).

, Cantor’s Theorem

A non trivial interval in R is uncountably infinite.

Addition in R

Let A, B ∈ R be cuts. Define

A + B := {a + b : a ∈ A, b ∈ B},

then,• A + B is a cut;• neutral element for addition: 0∗ := (−∞, 0);• negative of D-cut A: −A := {−b : b ∈ A↑} which is is also acut, where;

A↑ :=

{(a,∞) A = (−∞, a), a ∈ Q,Q \ A A irrational;

• A + (−A) = 0∗.

Positivity and order in R

A D-cut is positive if A > 0∗ (i.e. A % (−∞, 0) or 0 ∈ A).Let Rpos := {positive D-cuts},then,• ∀ A ∈ R, either A = 0∗, A ∈ Rpos or −A ∈ Rpos;• if A, B ∈ Rpos, then A + B ∈ Rpos;• let A, B ∈ R, then A < B iff B + (−A) ∈ Rpos.

Absolute value

The absolute value of the D-cut A is |A| ∈ Rpos defined by

|A| =

0 A = 0,

A A ∈ Rpos,

−A − A ∈ Rpos.

In particular (!) |(−∞, a)| = (−∞, |a|).

Multiplication

We first define multiplication on Rpos. Define the positive part ofthe positive D-cut A by A+ := A ∩ (0,∞), thenA = (−∞, 0] ∪ A+.For A, B ∈ Rpos, define

AB := (−∞, 0] ∪ {xy : x ∈ A+, y ∈ B+}

and for A, B ∈ R define

A · B :=

0 A = 0 or B = 0,

|A||B| A, B ∈ Rpos or − A, −B ∈ Rpos

−|A||B| else.

, Dedekind’s theorem

(R,+, ·) is a complete ordered field with respect to Rpos.

Archimedean ordered fields

The ordered field (F ,+, ·) is called archimedean if∀ x > 0, ∃ n ∈ N such that nx := x + · · ·+ x︸ ︷︷ ︸

n times

> 1;

(i.e. there are no ‘‘infinitesimals").

, A complete, ordered field (F ,+, ·) is archimedean.

Non-archimedean ordered field

Let (F,+, ·) be the field of rational functions} on R equippedwith regular addition and multiplication of functions.Define the “positive elements” of F by

Fpos :=

{R = PQ ∈ F : ∃ t > 0 so that Q(x) 6= 0 & R(x) > 0 ∀ x ∈ (0, t)}.

, Fpos is an ordering for (F,+, ·);

/ F is not archimedean with respect to Fpos.

The complex numbers (!Mהמרוכבי Mהמספרי)

Define the complex numbers by

C = R(√−1) := {x +

√−1y : x , y ∈ R},

with addition and multiplication to satisfy the normal laws ofarithmetic.

, (C,+·) is a field;

/ there is no ordering for (C,+·)

General existence of real roots

∀ a ∈ Rpos n ∈ N, ∃ ! n√a ∈ Rpos such that(n√

a)n = a.

Proof

n√a := LUB A where A := {x ∈ Rpos : xn < a}.

Limit of a sequence

Suppose bn ∈ R (n ∈ N). We say that

bn tends to (!Êל Pשוא) B ∈ R as n→∞

writtenbn → B ∈ R as n→∞; or bn −→

n→∞B

if∀ ε > 0, ∃ nε such that |bn − B| < ε ∀ n ≥ nε.

Example If an ≤ an+1 and {an : n ≥ 1} is bounded, thenan −→

n→∞LUB {an : n ≥ 1}.

Conditions for convergence

¶ Comparison

Suppose that an ≥ 0, an → 0 as n→∞ and thatM > 0, bn ∈ R, |bn| ≤ Man ∀ n ≥ 1, then bn → 0 as n→∞.

¶ Absolute value proposition

an → L as n→∞ iff |an − L| → 0 as n→∞ and in this case|an| → |L|.

¶ Sandwich principle

Suppose that an ≤ xn ≤ bn ∀ n ≥ 1 and that an → L, bn → L asn→∞, then xn −→

n→∞L.

Divergence to ∞

We say that the sequence (x1, x2, . . . ) diverges ( (מתבדרת! to ∞ (asn→∞) if for each M > 0, ∃ NM such that xn > M ∀ n ≥ NM

(and write this xn →∞).

, Let (x1, x2, . . . ) be an increasing sequence, then either(x1, x2, . . . ) is convergent, or xn →∞.

Examples

n∑k=1

1k −→n→∞

∞,n∑

k=0

12k−→n→∞

2.

Arithmetic of limits

Suppose that an → a and bn → b as n→∞, then

an + bn → a + b as n→∞; (1)

anbn → ab as n→∞; (2)

and in case b 6= 0:anbn→ a

bas n→∞; (3)

Accumulation points

Let E ⊂ R. A point x ∈ R is called an accumulation point of E if∀ ε > 0, # E ∩ (x − ε, x + ε) =∞.

, The following are equivalent for E ⊂ R and x ∈ R:

(i) x is an accumulation point of E ;

(ii) ∀ ε > 0, ∃ y ∈ E ∩ (x − ε, x + ε), y 6= x ;

(iii) ∃ (z1, z2, . . . ) ∈ EN such that zk 6= z` ∀ k 6= ` & zn −→n→∞

x .

- Bolzano-Weierstrass theorem (accumulation

points)

If E is an infinite, bounded set, then E ′ 6= ∅.The proof of the Bolzano-Weierstrass theorem uses:

, Cantor’s Lemma (or the Chinese box theorem)

A nested sequence of non-empty, closed intervals in R has anon-empty intersection.

Proof of the Bolzano-Weierstrass theorem

Suppose that E ⊂ I a closed, finite interval.For I = [a, b], write I− := [a, a+b

2 ] and I+ := [a+b2 , b];

I = I− ∪ I+ whence ∃ I1 = I± with # E ∩ I1 =∞;Similarly ∃ I2 = I±1 with # E ∩ I2 =∞;continuing, get closed intervals In ⊃ In+1 so that

• In+1 = I±n and # E ∩ In =∞ ∀ n ≥ 1.

Since |In| = |I |2n −→n→∞

0, by Cantor’s lemma,⋂∞

n=1 In = {Z} for

some Z ∈ R.

Z ∈ E ′. 2�

Limit points

Let E ⊂ R. A point x ∈ R is called a limit point ( גבול! (נקודת of E if∃ yn ∈ E (n ≥ 1) such that yn → x .

Closure ( (סגור! of E :

{limit points of E} =: E .

, E = E ∪ E ′

A set is closed if E = E .

- A closed subset of R which is bounded above (below) hasa maximal (minimal) element.

Subsequences

An integer subsequence ( !Mשלמי של (תתÊסידרה is an infinite subsetK ⊂ N, K = {n1, n2, . . . } arranged in increasing ordern1 < n2 < · · · → ∞.A subsequence of the sequence {a1, a2, . . . } is a sequence of form{an1 , an2 , . . . } where nk →∞ is an integer subsequence.

, Bolzano-Weierstrass Theorem (convergent

subsequences)

Every bounded sequence has a convergent subsequence.

Partial limits of a sequence

The partial limit set ( !Mהחלקיי גבולות (קבוצת PL(a1, a2, . . . ) of thebounded sequence (a1, a2, . . . ) is

PL(a1, a2, . . . ) := {a ∈ R : ∃ nk →∞, ank −→k→∞

a} 6= ∅.

, For (a1, a2, . . . ) a bounded sequence,

#PL(a1, a2, . . . ) = 1 ⇐⇒ ∃ limn→∞ an.

- Let (a1, a2, . . . ) be a bounded sequence, thenPL(a1, a2, . . . ) is closed and bounded.

Upper and lower limits

The upper limit (!Nעליו (גבול of the bounded sequence (a1, a2, . . . ) is

limn→∞

an := max PL(a1, a2, . . . )

and the lower limit ( !Nתחתו (גבול of the sequence (a1, a2, . . . ) is

limn→∞

an := min PL(a1, a2, . . . ).

, (a1, a2, . . . ) converges iff limn→∞ an = limn→∞ an.

- Let a = (a1, a2, . . . ) be a bounded sequence, then

(i) ∀ α < limn→∞ an, ∃ Nα such that an > α ∀ n > Nα;

(ii) ∀ β > limn→∞ an, K ≥ 1, ∃ N > K such that aN < β;

(i) ∀ ω > limn→∞ an, ∃ Nω such that an < ω ∀ n > Nω;

(ii) ∀ ξ < limn→∞ an, K ≥ 1, ∃ N > K such that aN > ξ;

Cauchy sequences

Or how to prove a sequence converges without knowing

the limit.

A sequence (a1, a2, . . . ) is called a Cauchy sequence if∀ ε > 0, ∃ Nε ≥ 1 such that

|an − an′ | < ε ∀ n, n′ ≥ Nε.

, Cauchy’s Theorem

A sequence converges ⇐⇒ it is a Cauchy sequence.

Newton’s approximation of√

2 ∈ R

i.e. construction of rational sequences yn −→n→∞

√2.

Given y > 0 define f (y) := y2 + 1

y ∈ R.

For y0 > 0, y20 > 2, define yn (n ≥ 1) by yn+1 := f (yn) (n ≥ 0).

It follows that yn > 0, y2n ≥ 2 ∀ n ≥ 0 and that

y0 ≥ y1 ≥ · · · ≥ yn ≥ yn+1 ≥ . . . .

The set A := {yn : n ≥ 0} is bounded below andyn → z := inf A =

√2.

Continued fractions

Let x ∈ (0, 1], then ∃ a ∈ N, X ∈ [0, 1) such that x = 1a+X .

¶1 Let x ∈ (0, 1), then x ∈ Q ⇐⇒ ∃ n ≥ 1, a1, . . . , an ∈ N suchthat

x =1||a1

+1||a2

+ · · ·+1||an

:=1

a1 + 1a2+...+ 1

an

.

¶2 Let x ∈ (0, 1) \Q, then∃ a1(x), . . . , an(x), · · · ∈ N & X1(x), . . . ,Xn(x), · · · ∈ (0, 1) \Qsuch that

x =1|

|a1(x)+

1||a2(x)

+· · ·+1|

|an(x)+Xn :=

1

a1(x) + 1a2(x)+...+ 1

an(x)+Xn

.

Continued fractions

¶3 For a ∈ N, x ∈ R define va(x) := 1a+x then for

a1, . . . , an ∈ N, x ∈ R

1||a1

+1||a2

+ · · ·+1|

|an + y= va1 ◦ · · · ◦ van(y) =: va1,...,an(y);

and |v ′a| ≤ 1, |v ′a,b| ≤14 . Whence |v ′a1,...,an | ≤

12n−1 and by MST

|va1,...,an(x)− va1,...,an(y)| ≤ 12n−1 ∀ x , y ∈ [0, 1].

Convergence of continued fractions

, Convergence of continued fractions

(i) ∀ x ∈ (0, 1) \Q,

1||a1(x)

+1|

|a2(x)+ · · ·+

1||an(x)

−→n→∞

x ;

(ii) if A1, . . . ,An, · · · ∈ N, then

∃ limn→∞

1||A1

+1||A2

+ · · ·+1||An

=: α ∈ [0, 1] \Q & An = an(α).

Proof ideas

Proof of (i)

|1|

|a1(x)+

1||a2(x)

+ · · ·+1|

|an(x)− x | = |va1,...,an(0)− va1,...,an(Xn)|

≤ 12n−1 .

(ii) sketch Write αn :=1||A1

+1||A2

+ · · ·+1||An

, then

|αn − αn+k | = |vA1,...,An(0)− vA1,...,An(vAn+1,...,An+k(0))| ≤ 1

2n−1.

Convergence of Averages

¶1 Convergence of arithmetic means

Suppose that xn −→n→∞

L, then

1

n

n∑k=1

xk −→n→∞

L.

¶2 Convergence of geometric means

Suppose that xn > 0 and xn −→n→∞

L > 0, then

(n∏

k=1

xk)1n −→

n→∞L.

Ratio theorem

, D’Alembert’s ratio theorem

Suppose that an > 0 (n ∈ N) and that an+1

an−→n→∞

L, then

a1nn −→

n→∞L.

- Corollaries

(i) n1n −→

n→∞1;

(ii)(2nn

) 1n −→

n→∞4.

Proposition e∃ lim

n→∞(n+1

n )n =: e ∈ (2, 3).

Series

The series∑∞

k=1 ak converges if∃ limn→∞

∑nk=1 ak =:

∑∞k=1 ak ∈ R.

, If∑∞

k=1 ak converges, then

(i) ∀ N ≥ 1 so does∑∞

k=N ak and

(ii)∑∞

k=N ak −→N→∞

0.

Series with non-negative terms

Write∑∞

k=1 ak <∞ for∑∞

k=1 ak converges; and∑∞k=1 ak =∞ for

∑∞k=1 ak diverges.

- Comparison of positive term series Suppose thatan, bn ≥ 0 (n ≥ 1) and that M > 0, N ≥ 1 are such thatan ≤ Mbn ∀ n ≥ N.If∑∞

n=1 bn <∞ , then∑∞

n=1 an <∞.

Absolute convergence of series (!Mטורי של בהחלט (התכנסותThe series

∑∞n=1 an is said to converge absolutely בהחלט!) מתכנס (הטור if∑∞

n=1 |an| <∞.

, If∑∞

n=1 |an| <∞ , then∑∞

n=1 an converges .

Proof (s1, s2, . . . ) is a Cauchy sequence where sn :=∑n

j=1 aj .

- Convergence of exponential series)

For x ∈ R, the series∑∞

k=0xk

k! converges absolutely;

(1 + xn )n −→

n→∞

∞∑n=0

xn

n! ∀ x ∈ R. (e)

whence e = limn→∞(1 + 1n )n =

∑∞n=0

1n! and

, e /∈ Q.

Proof

0 < e −q∑

j=0

1

j!≤ 4

9

1

q!.

Tests for convergence of series¶1 Cauchy’s Root test Suppose that an ≥ 0.

1) If lim supn→∞ a1nn < 1, then

∑∞n=1 an <∞.

2) If lim supn→∞ a1nn > 1, then an 9 0.

¶2 D’Alembert’s ratio theorem (lim version). Supposethat an > 0 for large n ∈ N.

(i) lim supn→∞ a1nn ≤ lim supn→∞

an+1

an;

(ii) lim infn→∞ a1nn ≥ lim infn→∞

an+1

an.

¶3 Cauchy’s condensation testSuppose that an ≥ an+1 ↓ 0 and let b ∈ N, b ≥ 2, then

∞∑n=1

an <∞ ⇔∞∑n=1

bnabn <∞.

¶4 Leibnitz’s Theorem Suppose that an ≥ an+1 ↓ 0, then theseries

∑∞n=1 an(−1)n+1 converges and 0 <

∑∞n=1 an(−1)n+1 < a1.

Power series and radius of convergence

Cauchy-Hadamard Theorem Let an ∈ R (n ≥ 0) and set

R :=1

lim n√|an|

∈ [0,∞].

a) If |x | < R, then the series∑∞

n=1 anxn converges absolutely,

andb) if |x | > R, then the series

∑∞n=1 anxn diverges.

The series∑∞

n=1 anxn is known as a power series and R is knownas its radius of convergence. It defines a R-valued function on(−R,R).

Real powers of positive real numbers

, Theorem exp ∃ an increasing bijection exp : R→ R+

so that exp(1) = e and

exp(x + y) = exp(x) exp(y) ∀ x , y ∈ R. (‡)

Theorem exp follows from

- Theorem log ∃ an increasing bijection log : R+ → R sothat log(e) = 1 and

log(xy) = log(x) + log(y) ∀ x , y ∈ R+. (†)

Powers and logsDefine for a > 0

ar := exp(r log(a)) ∀ r ∈ R.

, For a, b > 0, r , s ∈ R

(i) (ab)r = arbr , (ii) ar+s = aras , (iii) (ar )s = ars .

- Exponential continuity proposition

(i) If xn −→n→∞

t ∈ R, then exp(xn) −→n→∞

exp(t).

(ii) If an > 0 and an −→n→∞

a > 0, then log(an) −→n→∞

log(a).

Suppose that a > 0, t ∈ R and xn ∈ R.

(iii) If xn −→n→∞

t ∈ R, an > 0 and an −→n→∞

a > 0, then axnn −→n→∞at .

, Theorem e

ex =∞∑n=0

xn

n! ∀ x ∈ R.

Limits of functionsLet f : (a, b)→ R and suppose that A ⊂ (a, b), c ∈ A′.

We say that f (x)Heine−→

x→c, x∈AL if

xn ∈ A, xn → c =⇒ f (xn)→ L.

We say that f (x)Cauchy−→

x→c, x∈AL if

∀ ε > 0, ∃ δ > 0 such that |f (x)−L| < ε whenever x ∈ A, |x−c | < δ.

Equivalence TheoremLet f : (a, b)→ R and suppose that A ⊂ (a, b), c ∈ A′, then

f (x)Heine−→

x→c, x∈AL ⇐⇒ f (x)

Cauchy−→x→c, x∈A

L.

, (1 + x)1x −→

x→0, x∈R+

e.

Cauchy’s condition

Let A ⊂ (a, b) and suppose c ∈ A′. The function f : (a, b)→ R,satisfies Cauchy’s condition at c along A if

∀ ε > 0, ∃ δ > 0 such that |f (x)−f (y)| < ε whenever x ∈ (c−δ, c+δ)∩A.

Proposition For f : (a, b)→ R, A ⊂ (a, b) and c ∈ A′:

∃ limx→c, x∈A f (x) iff f : (a, b)→ R, satisfies Cauchy’s conditionat c along A.

Continuous functions

The function f : (a, b)→ R is continuous at c ∈ (a, b) iff (x) −→

x→cf (c);

The function f : (a, b)→ R is continuous on A ⊂ (a, b) if it iscontinuous at every c ∈ A.

The elementary functions (as in the list–rational,trigonometric, exponential, logarithmic functions) are continuouson their domains of definition.

Discontinuous examples

¶1The “salt and pepper function”Let D : R→ R be defined by

D(x) =

{1 x ∈ Q,0 x /∈ Q.

This function is continuous nowhere.

¶2The “salad cream function”e : R→ R defined by

e(x) =

0 x = 0,1q x = p

q , p ∈ Z \ {0}, q ∈ N, gcd(p, q) = 1,

0 x /∈ Q,

Proposition e is continuous at c iff c ∈ (R \Q) ∪ {0}.

Continuity of series

Theorem (General continuity of series)Suppose that un : [a, b]→ R are continuous, and that

∞∑n=1

supx∈[a,b]

|un(x)| <∞

then(i) the series U(t) :=

∑∞n=1 un(t) converges absolutely ∀ t ∈ [a, b];

(ii) the function U : [a, b]→ R is continuous.

Proof of (ii) ∀ N ≥ 1, t ∈ [a, b]

|U(t)− U(Z )| ≤ |N∑

n=1

un(t)−N∑

n=1

un(Z )|+ 2∞∑

n=N+1

supx∈[a,b]

|un(x)|.

EXAMPLES

¶1 Power series A power series is continuous on its open intervalof convergence.

¶2 Takagi-Rudin type functionsLet 〈x〉 := min {|x − 2n| : n ∈ Z}, then x 7→ 〈x〉 is continuous.

A Takagi-Rudin type function is a function of form

x 7→ Ta,d(x) =∞∑n=1

an〈dnx〉

where 0 < a < 1, d > 1For 0 < a < 1, d > 1, Ta,d is continuous on R.

¶3 Weierstrass’ functionsThe Weierstrass functions have form wa,b(x) :=

∑∞n=1

sin(bnπx)an for

a > 1 and b > 0. Again by the Continuity of series theorem,wa,b is continuous on R ∀ a > 1, b > 0.

Intermediate values

, Cauchy’s Intermediate value theorem (IVT) משפת)

!Mהבניי Kער)

The continuous image of an interval is an interval.

Onto propositionsThe following functions are bijections:(1) sin : [−π

2 ,π2 ]→ [−1, 1];

(2) cos : [0, π]→ [−1, 1];(3) tan : (−π

2 ,π2 )→ R.

Corollary Every polynomial of odd degree has a real zero.

IVT was needed in the construction of the Non-archimedean

ordered field.

Continuity of inverse and composition functions

TheoremSuppose that I ⊂ R is a bounded, closed interval and thatf : I → R is continuous and strictly monotone, then f : I → f (I ) isa bijection and f −1 : f (I )→ I is continuous.

CorollaryThe following functions are continuous:

• arcsin : [−1, 1]→ [−π2 ,

π2 ];

• arccos : [−1, 1]→ [0, π];

• arctan : R→ (−π2 ,

π2 ).

Theorem (Continuity of composition of functions)Suppose that I , J ⊂ R are intervals, and thatf : I → J, g : J → R. Let x ∈ I .If f is continuous at x and g is continuous at f (x), then g ◦ f iscontinuous at x.

Intermediate value property (IVP)Say that f : (a, b)→ R has the IVP if f (J) is an interval ∀intervals J ⊂ (a, b)

¶1 Let

f (x) =

{sin 1

x x 6= 0,

0 x = 0,

then f is continuous on R \ {0}; not continuous at 0, and has theIVP.

¶2 Here x =∑∞

n=1an2n ∈ [0, 1] in binary expansion.

(i) F : [0, 1]→ R defined by F (x) := limn→∞1n

∑nk=1 ak .

(ii) G : [0, 1]→ R defined by

G (x) :=

{limn→∞

∑nk=1

2ak−1k if this is finite;

0 else.

Both functions are continuous nowhere and have the IVP as theimage of every nontrivial interval is [0, 1] under F and R under G .

Monotone functions

¶1 one sided limits

Suppose that f : (a, b)→ R is monotone and that c ∈ (a, b), then

∃ limx→c±

f (x) =: f (c±).

¶2 continuity of monotone functions

Suppose that f : (a, b)→ R is monotone and that c ∈ (a, b),then f is continuous at c iff f (c−) = f (c+).

¶3 IVP of monotone functions

Suppose that I ⊂ R is a bounded interval and that f : I → R ismonotone, then f is continuous ⇔ f (I ) is an interval.

Monotone functions

¶4 continuity points of monotone functions

Suppose that I ⊂ R is an open interval and that f : I → R ismonotone, then {x ∈ I : f discontinuous at x} f is at mostcountable.

¶ExampleDefine f : (0, 1)→ R by

f (x) :=∞∑q=1

bqxc2q

where bxc = max {n ∈ Z : n ≤ x}.

Claim This function f is continuous at irrational points of (0, 1)and discontinuous at rational points of (0, 1).

Continuous functions on closed, bounded intervals

Weierstrass’ TheoremsSuppose that I ⊂ R is a interval, that f : I → R is continuous on Iand that A ⊂ I is a closed, bounded set, then f (A) is closed andbounded.

Consequently, f is bounded on A, and has maximal andminimal values.

Uniform continuityDefinitionSuppose that A ⊂ R and that f : A→ R. Say that f is uniformlycontinuous on A if∀ε > 0, ∃ δ > 0 such that

x , y ∈ A, |x − y | < δ =⇒ |f (x)− f (y)| < ε.

PropositionThe function f : A→ R is uniformly continuous on A iff

LUB {|f (x)− f (y)| : x , y ∈ A, |x − y | < t} −→t→0+

0.

Motivated by the above proposition, define the modulus ofcontinuity of a function f : A→ R by

ωf ,A(t) := LUB {|f (x)− f (y)| : x , y ∈ A, |x − y | < t} ≤ ∞where we set LUB A =∞ for A not bounded above.It follows (as above) that f : A→ R is uniformly continuous onA iff ωf ,A(t) −→

t→0+0.

The function f : A→ R is called Lipschitz continuous in case∃ M > 0 so that ωf ,A(t) ≤ Mt (t > 0).

Continuity on a closed, bounded set

¶1 Cantor’s Theorem

Suppose that A ⊂ R is closed and bounded. If f : A→ R iscontinuous on A, then uniformly continuous on A.

¶2 If J is a bounded interval then f : J → R is uniformlycontinuous ⇐⇒ f has a continuous extension to J.

Tangents and DifferentialsThe function f : (a, b)→ R is said to be differentiable at ( !Êב (גזירה

x ∈ (a, b) if ∃ f ′(x) ∈ R such that

f (x + h)− f (x)

h−→

h→0, h 6=0f ′(x).

The number f ′(x) ∈ R is known as the derivative (נגזרת!) of f at x .

¶ Routine Theorem on arithmetical operations

Suppose that u, v : (a, b)→ R are differentiable at c ∈ (a, b), thenu + v and uv are also differentiable at c with

(u + v)′(c) = u′(c) + v ′(c), (a)

(uv)′(c) = u′(c)v(c) + v ′(c)u(c).

In case v ′(c) 6= 0, uv is differentiable at c with

(uv )′(c) = u′(c)v(c)−u(c)v ′(c)v(c)2

. (b)

Examples that continuity ; differentiability

¶1 f (x) := |x |, x ∈ R.

¶2 f : R→ R defined by

f (x) =

{x sin 1

x x 6= 0,

0 x = 0.

• In both examples, f is differentiable at every x 6= 0.

A continuous, nowhere differentiable function

, Rudin’s function T (x) =∑∞

n=1(34)n〈4nx〉 iscontinuous, but nowhere differentiable where 〈x〉 is the minimumdistance to an even integer.

Proof sketch of non differentiability at Z ∈ R.

For N ∈ N define δN(Z ) = ± 12·4N so that there is no integer

strictly between 4NZ and 4N(Z + δN) = 4NZ ± 12 . This ensures

that

|〈4N(Z + δN)〉 − 〈4NZ 〉| =1

2

whence

|T (Z + δN)− T (Z )| > 3N

2|δN |.

Compositions and inverses

¶1 Derivative of composition (chain rule)

Suppose that I , J ⊂ R are open intervals and thatf : I → J, g : J → R.If f is differentiable at a ∈ I and g is differentiable at f (a), theng ◦ f is differentiable at a with

(g ◦ f )′(a) = g ′(f (a)) · f ′(a).

¶2 Derivative of inverse function

Let I = (a, b), let f : I → R be continuous and strictlymonotone and let f −1 : f (I )→ I be the inverse function.If x ∈ (a, b) and f is differentiable at x with f ′(x) 6= 0, then f −1

is differentiable at f (x) with f −1′(f (x)) = 1f ′(x) .

Complex valued functions

A complex function f = u +√−1v : (a, b)→ C is called

differentiable at x ∈ (a, b) if both its real and its imaginary partsu, v are differentiable at x . In this case

f ′(x) := u′(x) +√−1v ′(x) = lim

h→0, h 6=0

f (x + h)− f (x)

h.

For W ∈ C define EW : R→ C by EW (x) := exp[Wx ] whereexp[a +

√−1b] := ea cos b +

√−1ea sin b.

¶1 EW is differentiable on R and E ′W = WEW .

¶2 Corollary: differential equations and complex

numbers

Suppose that b0, . . . , bN ∈ R and that W = s +√−1t ∈ C

satisfies∑N

k=0 bkW k = 0, then∑N

k=0 bk f (k) ≡ 0 forf (x) = esx cos(tx), f (x) = esx sin(tx).

Example: a discontinuous derivativeDefine f : R→ R by

f (x) :=

{x2 cos 1

x x 6= 0

0 x = 0,

then

f ′(x) :=

{sin 1

x + 2x cos 1x x 6= 0

0 x = 0,

The derivative at x = 0 is calculated directly from the definition:

f (x)−f (0)x = x cos 1

x −→x→0, x 6=0

0 = f ′(0),

and the derivative at x 6= 0 is calculated using the chain rule.

Leibnitz’s theoremSuppose that u, v : (a, b)→ R are n-times differentiable on (a, b),then so is uv and

(uv)(n) =n∑

k=0

(n

k

)u(k)v (n−k).